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  Leetcode-041-缺失的第一个正数
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      <h1 id="Leecode-041-First-Missing-Positive"><a href="#Leecode-041-First-Missing-Positive" class="headerlink" title="Leecode-041-First Missing Positive"></a>Leecode-041-<a href="https://leetcode-cn.com/problems/first-missing-positive/" target="_blank" rel="noopener">First Missing Positive</a></h1><h2 id="思路：自哈希"><a href="#思路：自哈希" class="headerlink" title="思路：自哈希"></a>思路：自哈希</h2><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>给你一个未排序的整数数组，请你找出其中没有出现的最小的正整数。</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line">示例 1:</span><br><span class="line"></span><br><span class="line">输入: [1,2,0]</span><br><span class="line">输出: 3</span><br><span class="line">示例 2:</span><br><span class="line"></span><br><span class="line">输入: [3,4,-1,1]</span><br><span class="line">输出: 2</span><br><span class="line">示例 3:</span><br><span class="line"></span><br><span class="line">输入: [7,8,9,11,12]</span><br><span class="line">输出: 1</span><br><span class="line"></span><br><span class="line">&lt;!--more--&gt;</span><br></pre></td></tr></table></figure>

<p><strong>你的算法的时间复杂度应为O(<em>n</em>)，并且只能使用常数级别的额外空间。</strong></p>
<p><strong>做完本题，请继续做以下题目</strong></p>
<p>Leetcode–041 : <a href="https://leetcode-cn.com/problems/first-missing-positive/" target="_blank" rel="noopener">https://leetcode-cn.com/problems/first-missing-positive/</a></p>
<p>Leetcode–442 : <a href="https://leetcode-cn.com/problems/find-all-duplicates-in-an-array/" target="_blank" rel="noopener">https://leetcode-cn.com/problems/find-all-duplicates-in-an-array/</a></p>
<p>Leetcode–448 : <a href="https://leetcode-cn.com/problems/find-all-numbers-disappeared-in-an-array/" target="_blank" rel="noopener">https://leetcode-cn.com/problems/find-all-numbers-disappeared-in-an-array/</a></p>
<a id="more"></a>



<p><strong>Solution：</strong></p>
<p>参考题解：[leetcode-041][<a href="https://leetcode-cn.com/problems/first-missing-positive/solution/tong-pai-xu-python-dai-ma-by-liweiwei1419/]" target="_blank" rel="noopener">https://leetcode-cn.com/problems/first-missing-positive/solution/tong-pai-xu-python-dai-ma-by-liweiwei1419/]</a></p>
<p><strong>我们来循序渐进的看这个问题</strong></p>
<h2 id="方法一：哈希表"><a href="#方法一：哈希表" class="headerlink" title="方法一：哈希表"></a>方法一：哈希表</h2><p><strong>话在前头：此方法空间复杂度不符合要求</strong></p>
<ul>
<li>按照刚才读题的思路，其实我们只需要从最小的正整数1开始，依次判断2，3，4直到数组长度N是否在数组中。</li>
<li>如果当前考虑的数不在这个数组中，我们就找到了这个最小正整数。</li>
<li>由于我们需要依次判断某一个正整数是否在这个数组中，我们可以先把这个数组中所有的元素放进哈希表。接下来在遍历的时候，就能以O(1) 的时间复杂度判断某个正整数是否在这个数组中。</li>
<li>由于题目要求我们<strong>只能使用常数级别的空间</strong>，而哈希表的大小与数组的长度是线性相关的，因此空间复杂度不符合题目要求。</li>
</ul>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">firstMissingPositive</span><span class="params">(<span class="keyword">int</span>[] nums)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> len = nums.length;</span><br><span class="line"></span><br><span class="line">        Set&lt;Integer&gt; set = <span class="keyword">new</span> HashSet&lt;&gt;();</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 遍历一遍数组，全部都放进去</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> num : nums) &#123;</span><br><span class="line">            set.add(num);</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 再遍历一遍set，如果不存在就说明找到了那个正数</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt;= len; i++) &#123;</span><br><span class="line">            <span class="keyword">if</span> (!set.contains(i))&#123;</span><br><span class="line">                <span class="keyword">return</span> i;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 极端情况下是最后一个元素</span></span><br><span class="line">        <span class="keyword">return</span> len + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>



<ul>
<li><strong>时间复杂度</strong>：O(N)  这里 N 表示数组的长度。</li>
<li><strong>空间复杂度</strong>：O(N)  把 N个数存在哈希表里面，使用了 N个空间。</li>
</ul>
<h2 id="方法二：排序-二分查找"><a href="#方法二：排序-二分查找" class="headerlink" title="方法二：排序 + 二分查找"></a>方法二：排序 + 二分查找</h2><p><strong>话在前头：这个方法时间复杂度不符合要求</strong></p>
<ul>
<li>根据刚才的分析，这个问题其实就是要我们查找一个元素，而查找一个元素，如果是在有序数组中查找，会快一些；</li>
<li>因此我们可以将<strong>数组先排序</strong>，<strong>再使用二分查找</strong>法从最小的正整数 1 开始查找，找不到就返回这个正整数；</li>
<li>这个思路需要先对数组排序，而排序使用的时间复杂度是 O<em>(</em>N<em>log</em>N)，是不符合这个问题的时间复杂度要求。</li>
</ul>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span></span>&#123;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">firstMissingPositive</span><span class="params">(<span class="keyword">int</span>[] nums)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> len = nums.length;</span><br><span class="line">        <span class="comment">// 排序时间复杂度O(NlogN)</span></span><br><span class="line">        Arrays.sort(nums);</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 二分查找去查找每一个正数</span></span><br><span class="line">        <span class="comment">// 二分查找时间复杂度O（logn）</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt;= len; i++) &#123;</span><br><span class="line">            <span class="comment">// 一个一个去找</span></span><br><span class="line">            <span class="keyword">int</span> res = binarySearch(nums, i);</span><br><span class="line">            <span class="comment">// 如果没找到这个数，说明就是缺少的最小正整数</span></span><br><span class="line">            <span class="keyword">if</span> (res == -<span class="number">1</span>) &#123;</span><br><span class="line">                <span class="keyword">return</span> i;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> len + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">    <span class="comment">// 常用二分查找模板</span></span><br><span class="line">    <span class="function"><span class="keyword">private</span> <span class="keyword">int</span> <span class="title">binarySearch</span><span class="params">(<span class="keyword">int</span>[] nums, <span class="keyword">int</span> target)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> left = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span> right = nums.length - <span class="number">1</span>;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">while</span> (left &lt;= right) &#123;</span><br><span class="line">            <span class="comment">// JDK源码中就是这么写的</span></span><br><span class="line">            <span class="keyword">int</span> mid = (left + right) &gt;&gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span> (nums[mid] == target) &#123;</span><br><span class="line">                <span class="keyword">return</span> mid;</span><br><span class="line">            &#125; <span class="keyword">else</span> <span class="keyword">if</span> (nums[mid] &lt; target) &#123;</span><br><span class="line">                left = mid + <span class="number">1</span>;</span><br><span class="line">            &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">                right = mid - <span class="number">1</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> -<span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>



<h2 id="方法三：自哈希"><a href="#方法三：自哈希" class="headerlink" title="方法三：自哈希"></a>方法三：自哈希</h2><p><strong>所以自哈希 ：也就是将数组自身视为哈希表</strong></p>
<ul>
<li><p>我们可以<strong>把原始的数组当做哈希表来使用，事实上，哈希表本身其实也是一个数组。</strong></p>
</li>
<li><p>我们要找的数就在 <code>[1, N + 1]</code> 里，最后 <code>N + 1</code> 这个元素我们不用找。因为在前面的 <code>N</code> 个元素都找不到的情况下，我们才返回 <code>N + 1</code>；</p>
</li>
<li><p><strong>那么我们采用这样的思路</strong></p>
<ul>
<li>1 这个数 放到下标是0的位置</li>
<li>2 这个数 放到下标是1的位置</li>
<li>。。。</li>
<li>按照这个思路整理一遍数组</li>
</ul>
</li>
<li><p>然后我们再遍历一次数组，第一个遇到它的值不等于下标的那个数，就是我们要找的确实的第一个正数。</p>
</li>
<li><p><strong>这个思想就相当于我们自己编写哈希函数，这个哈希函数的规则特别简单，那就是数值是 i 的数映射到下标为 i-1 的位置</strong></p>
</li>
</ul>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span></span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">firstMissingPositive</span><span class="params">(<span class="keyword">int</span>[] nums)</span></span>&#123;</span><br><span class="line">        <span class="keyword">int</span> len = nums.length;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 对数组自己做哈希：数值为i的数字映射到下标 i - 1的位置</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; len; i++) &#123;</span><br><span class="line">            <span class="keyword">while</span> (nums[i] &gt; <span class="number">0</span> &amp;&amp; nums[i] &lt;= len &amp;&amp; nums[nums[i] - <span class="number">1</span>] != nums[i])&#123;</span><br><span class="line">                <span class="comment">// 满足在指定范围内，并且没有放在正确的位置上，才交换</span></span><br><span class="line">                <span class="comment">// 例如：数值3应该放在索引2的位置上</span></span><br><span class="line">                swap(nums,nums[i] - <span class="number">1</span>,i);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 遍历数组找到缺失的最小正数</span></span><br><span class="line">        <span class="comment">// 缺失的正整数是下标 + 1（i 从0 开始）</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; len; i++) &#123;</span><br><span class="line">            <span class="keyword">if</span> (nums[i] != i + <span class="number">1</span>)&#123;</span><br><span class="line">                <span class="keyword">return</span> i + <span class="number">1</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 都正确范围数组长度 + 1</span></span><br><span class="line">        <span class="keyword">return</span> len + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="comment">// 交换函数</span></span><br><span class="line">    <span class="function"><span class="keyword">private</span> <span class="keyword">void</span> <span class="title">swap</span><span class="params">(<span class="keyword">int</span>[] nums, <span class="keyword">int</span> i, <span class="keyword">int</span> j)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> temp = nums[i];</span><br><span class="line">        nums[i] = nums[j];</span><br><span class="line">        nums[j] = temp;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>



<p><strong>复杂度分析</strong></p>
<ul>
<li><strong>时间复杂度：O(n)</strong><ul>
<li><code>while</code>循环不会每一次都把数组里面的所有元素过一遍。如果有一些元素在这一次循环中被交换到他们应该在的位置上，那么在后序的遍历中，由于他们已经在正确的位置上了，代码执行到他们的时候，就会被跳过（<strong>相当于是一个剪枝的处理）</strong></li>
<li>最极端的一种情况是，在第 1 个位置经过这个 while 就把所有的元素都看了一遍，这个所有的元素都被放置在它们应该在的位置，那么 for 循环后面的部分的 while 的循环体都不会被执行。</li>
<li>平均下来，每次元素只要被看一次就可以了。<code>while</code> 循环体被执行很多次的情况不会发生。这样的复杂度分析的方法叫做<strong>均摊复杂度分析。</strong></li>
</ul>
</li>
<li><strong>空间复杂度：O(1)</strong> </li>
</ul>

      
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